## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 11 - Section 11.5 - Lines and Planes in Space - Exercises - Page 631: 53

#### Answer

$(\dfrac{3}{2},\dfrac{-3}{2},\dfrac{1}{2})$

#### Work Step by Step

Parametric equations are: $x=1-t; y=3t; z=1+t$ We have the equation of the plane: $2x-y+3z=6$ Now, we plug in all of the parametric equations into the above equation of a plane: $t=\dfrac{-1}{2}$ Thus, our parametric equations become: $x=1-(\dfrac{-1}{2})=\dfrac{3}{2}; y=3(\dfrac{-1}{2})=\dfrac{-3}{2}; z=1+(\dfrac{-1}{2})=\dfrac{1}{2}$ Hence, the line will meet at the point: $(\dfrac{3}{2},\dfrac{-3}{2},\dfrac{1}{2})$

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